My Favorite Things
About Financial Math!!!
1). Simple Interest – When money is
borrowed or lent, the borrower usually pays the lender for the service. The
amount charged is usually called interest.
Formula: I=PRT/100
·
I= interest
·
P= principal amount
·
T= time
·
R= rate per annum
Practice
Problem: If $600 is invested in an account and
earns $136.50 simple interest over a 3.5 year period, find the annual interest
rate.
·
Unknown = R
·
I = PRT/100 changes to: R =100xI/PT
·
R = 100 x 136.50/600 x 3.5 = 6.5%
2). Compound
Interest – It is
common for investors to add (compound) the interest to the principle so
that the principle grows during the term of the investment.
·
A= amount of the account
·
P= principal amount
·
T= time
·
R= rate
·
N= # of compounding
periods
Practice
Problem: $1200 is placed in a savings account that
pays 8% annual interest compounded annually. Find the amount in the account and
the interest paid after 10 years.
·
P = 1200, R = 8, n = 10
·
Formula: A = 1200(1 + 8/100)^10 = 2590.71
·
Account will contain $2590.71 after 10
years.
o
The interest is $2590.71 - $1200 =
1390.71Interest rate is annual yet it is calculated monthly! Express the
interest rate as % per month by dividing it by 12 to get R = 0.5%. The number
of interest periods must also be expressed in months (n = 60 months)
·
A = 1500(1 + 0.5/100)^60 = 2023.2752 or 2023 marks
·
The interest paid is:
o
2023 – 1500 = 523 marks
3). Currency Conversions
Practice Problem 1:
US
|
Aus
|
Sterling
|
Yen
|
Mark
|
1
|
1.633
|
0.610
|
139.490
|
1.782
|
·
Convert:
o $50 US to Japanese Yen
§
$1 US = 139.490 Yen
§
$50 US x 139.490 =
6974.5 Yen
o 15 Sterling to $US
§
$1 US = 0.61 Sterling
§
1 Sterling = 1/0.61 =
1.6393443 dollars
§
15 Sterling =x 1.6393443
= 24.59016 (or $24.59)
·
Alternative: 15/0.61 =
24.59016 (or $24.59)
o $1500 Australian to Marks
§
$1.633 Australian =
1.782 Marks
§
$1 = 1.782/1.633 =
1.0912431
§
1500 = 1.0912431 =
1636.86 Marks
·
Alternative: 1.782/1.633
x 1500 = 1636.86
o 125 Yen to Sterling
§
1 Yen in Sterling =
0.610/139.490 = 0.00437307
§
125 Yen = 0.00437307 x
125 = 0.5466
§
Alternative:
0.610/139.490 x 125 = 0.5466 (roughly 0.56 Sterling)
Practice
Problem 2:
|
GBP
|
USD
|
AUD
|
GBP
|
1
|
1.63
|
2.5
|
USD
|
P
|
1
|
Q
|
AUD
|
1/2.5
|
.61
|
1
|
-750 USD x AUD: (without commission)
-750 x 1/.61 = $1,229.5 AUD
How many AUD do you have if you take 2% commission?
-750 x .02 = 15
(750-15=735)
-735 x (1/.61) = $1,204.5
AUD
4). Construction and Use of Tables- Buy and Sell
Practice
Problem:
A.) 500 AUD- Simple Interest= 7% for 5 years
B.) 500 AUD to GBP
-Sell at 105 & Buy
at 100
Convert 7.5 compound quarterly for 5 years
A.) I=PRT/100 --> I=(500)(7)(5)/100 --> I=175
B.) 500 x 105 =52,500(1+((7.5/4)/100)^(4x5)
=76122/100
GBP --> =761 in AUD
Therefore…
A.) 500+175= 675 AUD
B.) 761 AUD
Thus...Option "B" is the best option!
5). Linear Programming – Commonly used
method in business management in making decisions on operating procedures.
·
Stage 1- Define the
variables
·
Stage 2- Write the constraints
of the problem as inequalities
·
Stage 3- Find the
objective function
·
Stage 4- Plot graphs of
the inequalities and identify the feasible region for the problem
·
Stage 5- Use the
objective function to find the optimal solution to the problem
Practice Problem:
1 metre of power cable costs $2 and 1 metre of coaxial cable costs $3. Menara
has $60 to spend. Write the information as inequalities and show the feasible
region as a graph. What is the largest amount of cable the Menara can afford to
buy?
·
Stage 1- What are you
asked to find? The amount of each type of cable that Menara can afford to buy.
o
P = Number of metres of
power cable that Menara buys
o
C = Number of metres of
coaxial cable
·
Stage 2- Menara cannot
buy a negative amount of either type of cable
o P is (greater than
or equal to) 0, and C is (greater than or equal to) 0
o 2p + 3c is
(less than or equal to) 60
·
Stage 3- What is the
objective function? Menara wants to buy as much cable of either type as
possible
o
Mathematically: Menara
wants to maximize A = p + c (the total length of cable bought it p+c)
·
Stage 4- (Due to web
page construction constraints, the feasible region graph cannot be drawn)
·
Stage 5- Find the
solution that gives the largest value
o
Vertices: p=0, c=0, A=0
(origin)/ p=30, c=0, A=30/ p=0, c=20, A=20
·
Largest value of the
objective = 30…thus, the full solution is: Menara should buy 30 metres of power
cable, which will cost him $60.
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Created by: Caitlin Doyle, Sarah
Johnson, and Hedieh Fakhriyazdi (P.1- Pandza)